Problem: If $m+\frac{1}{m}=8$, then what is the value of $m^{2}+\frac{1}{m^{2}}+4$?
Squaring the equation provided, we get $m^2+2(m)\left(\frac{1}{m}\right) +\frac{1}{m^2}=64,$ so $m^2+\frac{1}{m^2}+4=\boxed{66}$.